Let `4x^(2)-4(alpha-2)x + alpha-2=0(alpha in R)` be a quadratic equation. Find the values of `alpha` for which
One root is greater than `(1)/(2)` and other is smaller than 0

Answer

To solve the problem, we need to analyze the quadratic equation given by:

\[ 4x^2 - 4(\alpha - 2)x + (\alpha - 2) = 0 \]

We want to find the values of \( \alpha \) such that one root is greater than \( \frac{1}{2} \) and the other root is smaller than \( 0 \).

### Step 1: Identify the coefficients
The coefficients of the quadratic equation are:
- \( a = 4 \)
- \( b = -4(\alpha - 2) = -4\alpha + 8 \)
- \( c = \alpha - 2 \)

### Step 2: Use the quadratic formula
The roots of the quadratic equation can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{D}}{2a} \]

where \( D \) is the discriminant given by \( D = b^2 - 4ac \).

### Step 3: Calculate the discriminant
We need to ensure that the discriminant \( D \) is greater than \( 0 \) for the quadratic to have real roots:

\[ D = (-4\alpha + 8)^2 - 4 \cdot 4 \cdot (\alpha - 2) \]

Calculating \( D \):

\[ D = (16\alpha^2 - 64\alpha + 64) - 16(\alpha - 2) \]
\[ D = 16\alpha^2 - 64\alpha + 64 - 16\alpha + 32 \]
\[ D = 16\alpha^2 - 80\alpha + 96 \]

Setting \( D > 0 \):

\[ 16\alpha^2 - 80\alpha + 96 > 0 \]

### Step 4: Solve the inequality
To find the roots of the quadratic equation \( 16\alpha^2 - 80\alpha + 96 = 0 \), we can use the quadratic formula:

\[ \alpha = \frac{-b \pm \sqrt{D}}{2a} = \frac{80 \pm \sqrt{(-80)^2 - 4 \cdot 16 \cdot 96}}{2 \cdot 16} \]
\[ = \frac{80 \pm \sqrt{6400 - 6144}}{32} \]
\[ = \frac{80 \pm \sqrt{256}}{32} \]
\[ = \frac{80 \pm 16}{32} \]

Calculating the roots:

\[ \alpha_1 = \frac{96}{32} = 3, \quad \alpha_2 = \frac{64}{32} = 2 \]

### Step 5: Determine the intervals
The roots divide the number line into intervals. We need to test the intervals \( (-\infty, 2) \), \( (2, 3) \), and \( (3, \infty) \) to see where \( 16\alpha^2 - 80\alpha + 96 > 0 \).

1. For \( \alpha < 2 \): Choose \( \alpha = 0 \)
\[ 16(0)^2 - 80(0) + 96 = 96 > 0 \]

2. For \( 2 < \alpha < 3 \): Choose \( \alpha = 2.5 \)
\[ 16(2.5)^2 - 80(2.5) + 96 = 16(6.25) - 200 + 96 = 100 - 200 + 96 = -4 < 0 \]

3. For \( \alpha > 3 \): Choose \( \alpha = 4 \)
\[ 16(4)^2 - 80(4) + 96 = 256 - 320 + 96 = 32 > 0 \]

Thus, the intervals where \( D > 0 \) are \( (-\infty, 2) \) and \( (3, \infty) \).

### Step 6: Check conditions for roots
Next, we need to check the conditions for the roots:

1. **Condition 1**: One root must be greater than \( \frac{1}{2} \).
2. **Condition 2**: The other root must be less than \( 0 \).

Using Vieta's formulas, the sum of the roots \( r_1 + r_2 = -\frac{b}{a} = \frac{4\alpha - 8}{4} = \alpha - 2 \) and the product of the roots \( r_1 r_2 = \frac{c}{a} = \frac{\alpha - 2}{4} \).

### Step 7: Analyze the roots
To satisfy the conditions:
- \( r_1 > \frac{1}{2} \) and \( r_2 < 0 \) implies \( r_1 + r_2 < \frac{1}{2} \) (since one root is positive and the other is negative).

Thus, we need:

\[ \alpha - 2 < \frac{1}{2} \implies \alpha < \frac{5}{2} \]

### Step 8: Combine conditions
Now we combine the conditions:
- From the discriminant: \( \alpha < 2 \) or \( \alpha > 3 \)
- From the roots condition: \( \alpha < \frac{5}{2} \)

The only feasible solution is:

\[ \alpha < 2 \]

### Conclusion
The values of \( \alpha \) for which one root is greater than \( \frac{1}{2} \) and the other is smaller than \( 0 \) are:

\[ \alpha \in (-\infty, 2) \]